Inelastic collisions with constant momentum

In summary, the conversation discusses the calculation of kinetic energy before a collision and the discrepancy of the required answer of 1/2 mv². The expert agrees with the calculation and notes a potential error in the book. It is mentioned that the books specifies "identical" objects, but it is clarified that objects of the same mass are all that is required kinematically. The use of "identical" may be for economy of words and to convey the idea of symmetry. However, it is pointed out that the book actually states "the same two objects", implying that they are the same as in the preceding question.
  • #1
haha0p1
46
9
Homework Statement
The total momentum before the collision in an inelastic collisions is 0, but the total kinetic energy before the collision is 1/2mv². Calculate how the total kinetic energy before collision is 1/2mv².
Relevant Equations
Ek=1/2mv²
Kinetic energy before collision =1/2 mv² + 1/2 mv² = mv² (since energy is a scalar quantity, the direction does not matter). Kindly tell why am I not getting the required answer i.e: 1/2 mv². Am I doing the calculation wrong?
IMG_20230102_154827.jpg
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Hi,

I agree with your calculation. An unfortunate error in the book.

##\ ##
 
  • Like
Likes haha0p1
  • #3
It's interesting that the book specifies the "same object" or "identical" objects, where all that is required kinematically is objects of the same mass!
 
  • #4
PeroK said:
It's interesting that the book specifies the "same object" or "identical" objects, where all that is required kinematically is objects of the same mass!
Truee
 
  • #5
PeroK said:
It's interesting that the book specifies the "same object" or "identical" objects, where all that is required kinematically is objects of the same mass!
I think it's economy of words. "Identical objects" is shorter than "objects of the same mass" and conveys the idea of symmetry. "Identical" becomes relatively conciser when the masses also carry equal charges.
 
  • Skeptical
Likes PeroK
  • #6
PeroK said:
It's interesting that the book specifies the "same object"
No, it says "the same two objects". Presumably the same two as in the preceding question.
 

FAQ: Inelastic collisions with constant momentum

What is an inelastic collision?

An inelastic collision is a type of collision where the colliding objects do not bounce away from each other completely. Instead, a portion of the kinetic energy is converted into other forms of energy, such as heat or sound, and the objects may stick together or deform.

How is momentum conserved in an inelastic collision?

In an inelastic collision, the total momentum of the system is conserved. This means that the sum of the momenta of the colliding objects before the collision is equal to the sum of their momenta after the collision. Mathematically, this can be expressed as: \( m_1 \cdot v_1 + m_2 \cdot v_2 = (m_1 + m_2) \cdot v_f \), where \( m_1 \) and \( m_2 \) are the masses and \( v_1 \), \( v_2 \), and \( v_f \) are the velocities of the objects before and after the collision.

What is the difference between elastic and inelastic collisions?

In an elastic collision, both momentum and kinetic energy are conserved, meaning the total kinetic energy of the system remains the same before and after the collision. In contrast, in an inelastic collision, only momentum is conserved, while some of the kinetic energy is transformed into other forms of energy.

Can you provide an example of an inelastic collision?

One common example of an inelastic collision is a car crash. When two cars collide, they often crumple and stick together, converting some of their kinetic energy into deformation, heat, and sound. Despite this loss of kinetic energy, the total momentum of the system (both cars) remains conserved.

How do you calculate the final velocity in a perfectly inelastic collision?

In a perfectly inelastic collision, the colliding objects stick together after the collision. The final velocity can be calculated using the conservation of momentum. If \( m_1 \) and \( m_2 \) are the masses of the two objects and \( v_1 \) and \( v_2 \) are their initial velocities, the final velocity \( v_f \) is given by: \( v_f = \frac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_1 + m_2} \).

Similar threads

Why is momentum conservation always true in a collision?
Replies
4
Views
1K
Force and change in momentum in an inelastic/elastic collision
Replies
19
Views
2K
Why momentum of a ball bounced off a wall increases twice fold?
Replies
9
Views
3K
Linear Motion and Linear Momentum
Replies
31
Views
3K
Elastic Collision and Momentum of Ice Skaters
Replies
16
Views
3K
Potential Energy - 2D inelastic collision - ballistic pendulum
Replies
10
Views
2K
A planet of mass M and an object of mass m
Replies
6
Views
996
Hinged rod rotating, falling and hitting a mass
Replies
11
Views
2K
Impulse and perfectly inelastic collision between 2 points
Replies
4
Views
389
Elastic Collision of Hydrogen and Carbon Atoms
Replies
9
Views
2K

Hot Threads

Recent Insights

Back
Top